Princetonboy

Marathon Match 37

Sun, 20 Jul 2008 15:07:11 GMT

这个双休日没有出去玩也没有好好休息,牺牲了学习计划的时间,在家里做MM37. 这道题目的大致意思是: 给定一个有两种纹理的图片,要求分割出这两种纹理,一种纹理用0填写,另一种用1填写. 记分方式是判断正确的点数越多则分数越高. 题目告诉了纹理的生成的方法,以及给定的图片的生成方法. 这里图片的像素点全部用0或者1表示黑或者白.

纹理的生成方法如下: 给定一张随机的正方形图片,随机的意思是图片中的像素点值要么是0要么是1. 然后做如下操作.

To generate a texture, the number of axes is chosen between 2 and 4, inclusive, each being vertical (dx=0, dy=1), horizontal (dx=1, dy=0) or diagonal (dx=dy=1 or dx=-dy=1). Several axes of the set can have the same direction. For each axis, its "weight" is chosen between 1 and 5.

The neighborhood of a pixel (x,y) for a chosen set of axes is generated as follows:

pixel (x,y) is added to the neighborhood

for each axis

for k = 1 .. weight

pixels (x+dx*k,y+dy*k) and (x-dx*k,y-dy*k) are added to the neighborhood

The initial state of the texture is generated randomly by assigning each pixel color '0' or '1'.

The procedure of adjusting a pixel's color is as follows:

if the number of '1's in its neighborhood is strictly greater than the number of '0's

set the color of the pixel to '1'

else

set the color of the pixel to '0'

The image is then modified iteratively:

for k = 1 .. 2*S*S

pick random pixel

adjust its color

for i = 1 .. S

for j = 1 .. S

pick pixel (i,j)

adjust its color

Finally, the image is inverted with a probability of 1/3.

给定图片的生成方法如下: 给定了一张图片,随机的在上面找到四个点,其中两个点涂成黑色(用1表示),另外两个点涂成白色(用0表示). 然后用如下方法生成该图片.

To generate an image with two textures, two images are generated independently, as above. We then use a stochastic flood fill algorithm to determine which pixels of the final image are colored according to which texture. Four (unique) pixels out of an SxS pixel image are chosen randomly. Two of these are colored black, and two are colored blue. We then perform the following to color the rest of the image, where neighbor means adjacent in one of the four cardinal directions:

while(there is an uncolored pixel)

pick a colored pixel (i,j) with an uncolored neighbor (ii,jj)

color the (ii,jj) with the color of (i,j)

If this procedure produces more than two colored regions (a colored region is a connected component of all the same color), it is repeated from scratch.

Finally, the black pixels are replaced with the pixel values from one texture, and the blue pixels are replaced with the pixel values from the second texture.

这道题目的难点在于纹理图片与给定的图片均是通过随机的方法生成的,我们很难确定哪些点是属于哪个纹理的. 而且纹理是随机生成的,我们也不可能枚举出所有的可能的纹理图片,因此这道题目的解法也应该是用随机算法来解决的.

其实到目前为止我没有很好的想法,自己到目前为止的版本如下:

Highest Score : 34235.69

Lowest Score : 33227.00

#include <iostream>

#include <string>

#include <vector>

#include <ctime>

#include

using namespace std;

#define MAX 300

int which[MAX][MAX] ;

int qx[MAX*MAX] ;

int qy[MAX*MAX] ;

boolb[MAX][MAX] ;

const int dx[] = {1,0,-1,0} ;

const int dy[] = {0,1,0,-1} ;

class Textures

{

public:

void ff(int i, int j, int Size)

{

int h = 0, t = 1 ;

qx[0] = i ;

qy[0] = j ;

while(h < t)

{

i = qx[h] ;

j = qy[h] ;

h ++ ;

for(int k = 0 ; k < 4 ; k ++)

{

int ii = i+dx[k] ;

int jj = j+dy[k] ;

if(ii>=0&&ii<Size&&jj>=0&&jj<Size&&which[ii][jj] == which[i][j]&&!b[ii][jj])

{

b[ii][jj] = true ;

qx[t] = ii ;

qy[t] = jj ;

t++ ;

}

void segment(int Size)

{

memset(which,0,sizeof(which)) ;

int* lx = new int[Size*Size] ;

int* ly = new int[Size*Size] ;

int cnt = 4;

int x1, y1, i, j ;

for(i = 0 ; i < cnt ; i ++)

{

do{

x1 = rand()%(Size-1) + 1 ;

y1 = rand()%(Size-1) + 1 ;

}while(which[x1][y1] != 0) ;

lx[i] = x1 ;

ly[i] = y1 ;

which[x1][y1] = i%2 + 1 ;

}

int colored = cnt ;

while(colored < Size*Size)

{

int ptr = rand() % cnt ;

int x = lx[ptr] ;

int y = ly[ptr] ;

int n = 0 ;

for(i = 0 ; i < 4 ; i ++)

{

int xx = x + dx[i] ;

int yy = y + dy[i] ;

if(xx<0||yy<0||xx==Size||yy==Size) continue ;

if(which[xx][yy] == 0) n ++ ;

}

if(n == 0)

{

lx[ptr] = lx[--cnt] ;

ly[ptr] = ly[cnt] ;

continue ;

}

int xx = x ;

int yy = y ;

while(true)

{

x = xx ;

y = yy ;

int d = rand() % 4 ;

if(d == 0) x ++ ;

else if(d == 1) y ++ ;

else if(d == 2) x -- ;

else if(d == 3) y -- ;

if(x<0||y<0||x==Size||y==Size) continue ;

if(which[x][y] != 0) continue ;

which[x][y] = which[xx][yy] ;

colored ++ ;

lx[cnt] = x ;

ly[cnt] = y ;

cnt ++ ;

break ;

}

memset(b,false,sizeof(b)) ;

int regions = 0 ;

for(i = 0 ; i < Size ; i ++)

{

for(j = 0 ; j < Size ; j ++)

{

if(!b[i][j])

{

ff(i, j, Size) ;

regions ++ ;

}

if(regions > 2)

{

segment(Size) ;

}

for(i = 0 ; i < Size ; i ++)

{

for(j = 0 ; j < Size ; j ++)

{

if(which[i][j] != 1)

which[i][j] = 0 ;

}

vector <string> recognize(vector <string> image)

{

int Size = image.size() ;

vector <string> res ;

stringstr ;

srand((unsigned)time(NULL)) ;

segment(Size) ;

for (int i = 0 ; i < Size ; i ++)

{

str = "" ;

for (int j = 0 ; j < Size ; j ++)

{

str += which[i][j] + '0' ;

}

res.push_back(str) ;

}

return res ;

}

};

...

SRM407 Division II

Thu, 17 Jul 2008 15:07:57 GMT

Division II of Level One : SpiralWalking

Categores : Simulation

#include <iostream>

#include <string>

#include <vector>

#include

using namespace std ;

#define MAX 50

int val[MAX][MAX] ;

class SpiralWalking

{

public:

int totalPoints(vector <string> levelMap)

{

int result = 0 ;

int n = levelMap.size() ;

int m = levelMap[0].size() ;

memset(val,0,sizeof(val)) ;

for (int i = 0 ; i < n ; i ++)

{

for (int j = 0 ; j < m ; j ++)

{

val[i][j] = char(levelMap[i][j]) - 48 ;

result += val[i][j] ;

}

int x1 = 0 , y1 = m-1 ;

int x2 = n-1 , y2 = m-1 ;

int x3 = n-1 , y3 = 0 ;

int x4 = 1 , y4 = 0 ;

while (true)

{

if((x1 == x2 && y1 == y2) || (x1 == x3 && y1 == y3)

|| (x1 == x4 && y1 == y4))

break ;

else

result -= val[x1++][y1--] ;

if ((x2 == x1&& y2 == y1) || (x2 == x3 && y2 == y3)

|| (x2 == x4 && y2 == y4))

break ;

else

result -= val[x2--][y2--] ;

if ((x3 == x1 && y3 == y1) || (x3 == x2 && y3 == y2)

|| (x3 == x4 && y3 == y4))

break ;

else

result -= val[x3--][y3++] ;

if ((x4 == x1 && y4 == y1) || (x4 == x2 && y4 == y2)

|| (x4 == x3 && y4 == y3))

break ;

else

result -= val[x4++][y4++] ;

}

return result ;

}

};

Division II of Level Two & Division I of Level One : CorporationSalary

Categores : Graph Theory , Search

/* 用递归方法竟然开销过大,超时了! */

#include <iostream>

#include <string>

#include <vector>

using namespace std ;

class CorporationSalary

{

public:

long long res(vector <string> str, int m)

{

long long result = 0 ;

for (int i = 0 ; i < str[m].size() ; i ++)

{

if (str[m][i] == 'Y')

{

result += res(str, i) ;

}

if (!result)

return 1 ;

else

return result ;

}

long long totalSalary(vector <string> relations)

{

long long result = 0 ;

for (int i = 0 ; i < relations.size() ; i ++)

{

result += res(relations,i) ;

}

return result ;

}

};

/* 改写为用搜索方法,当n为MAX时,不到2ms过了! */

#include <iostream>

#include <string>

#include

#include <vector>

using namespace std ;

#define MAX 50

long long result[MAX] ;

boolflag[MAX] ;

class CorporationSalary

{

public:

long long totalSalary(vector <string> relations)

{

int i , j ;

long long sum = 0 ;

int n = relations.size() ;

memset(result,0,sizeof(result)) ;

memset(flag,false,sizeof(flag)) ;

while (true)

{

i = 0 , j = 0 ;

while (flag[i])

{

i ++ ;

}

if (i == n)

{

break ;

}

for (i = 0 ; i < n ; i ++)

{

if (!flag[i])

{

for (j = 0 ; j < n ; j ++)

{

if (relations[i][j] == 'Y' && flag[j] == true)

{

result[i] += result[j] ;

}

else if (relations[i][j] == 'Y' && flag[j] == false)

{

result[i] = 0 ;

break ;

}

if (!result[i] && j == n)

{

result[i] = 1 ;

flag[i] = true ;

}

else if (result[i] && j == n)

{

flag[i] = true ;

}

sum = 0 ;

for (i = 0 ; i < n ; i ++)

{

sum += result[i] ;

}

return sum ;

}

};

...

2008.07.14-2008.07.20工作学习计划

Sun, 13 Jul 2008 10:07:19 GMT

2008.07.14-2008.07.20工作学习计划 :

1. 数学 : 高等数学第一章;

2. 英语 : 熟悉单词A部分;

3. 数据结构 : 数据结构算法与应用-C++语言描述第一章;

4. 算法设计研究 : 动态规划;

5. 系统分析师考试 : 教程中"多媒体基础知识"的内容;

6. 工作 : NR600E上的E1板的开发.

...